Chapter 13 - Solutions

= Chapter 13 - Solutions Description: Find the weight of a cylindrical iron rod given its area and length and the density of iron. Part A On a part-time job you are asked to bring a cylindrical iron rod of density 7800 length 92.1 and diameter 2.75 from a storage room to a machinist. Calculate the weight of the rod. The acceleration due to gravity = 9.81. A.1 General approach To calculate the weight of a body you need to know its mass. If the mass is unknown and instead the problem gives you information on the volume of the body and the density of the material of which it is made you can calculate the mass of that body by applying the definition of density. A.2 Weight of a body A body of mass has weight with magnitude given by where = 9.81 is the acceleration due to gravity. A.3 Find the mass of the rod From the length and the cross-sectional area of the rod and the density of iron find the mass the rod. A.3.1 Finding mass from volume and density of Consider a mass of homogeneous material whose volume is. The density of that material is defined as. A.3.2 Volume of a cylinder The volume of a cylinder of length and cross-sectional area is given by If the cross section is a circle with radius then Express your answer numerically in kilograms.

If the cross section is a circle with radius then Express your answer numerically in kilograms. = Express your answer numerically in newtons. = Part B Now that you know the weight of the rod do you think that you will be able to carry the rod without a cart? B.1 Weight and mass A weight measured in newtons may not be very familiar to you; therefore it is more convenient to determine the corresponding mass measured in kilograms or in pounds. B.2 Unit conversion factors The unit convertion factors for newtons kilograms and pounds are. yes no ± Volume of Copper Description: ± Includes Math Remediation. Calculate the volume of a given number of moles of copper. Part A

The weight of a column of seawater 1 in cross section and 10 high What is the volume of a sample of 3.70 of copper? The atomic mass of copper (Cu) is 63.5 A.1 and the density of copper is. How to approach the problem Since the number of moles of copper is known calculate its mass and then use the mass and the density of copper to find the volume of the copper. A.2 What is the mass of the copper? Calculate the mass of 3.70 of copper. Express your answer in grams. = A.3 Using the density in calculations Recall that density is defined as the ratio of the mass to volume: the units since the volume should be expressed in cubic centimeters. Express your answer in cubic centimeters to 3 significant figures.. Be careful with = Pressure in the Ocean Description: Short conceptual problem on hydrostatic pressure at various depths in the ocean. This problem is based on Young/Geller Conceptual Analysis 13.2 The pressure at 10 below the surface of the ocean is about 2.00 10 5. Part A Which of the following statements is true? A.1 How to approach the problem Since pressure is defined as force per unit area the pressure at a given depth below the surface of the ocean is the normal force on a small area at that depth divided by that area. Thus given the pressure at 10 you can determine the normal force on a unit area at that depth. Also recall that in general the pressure in a fluid varies with height but it also depends on the external pressure applied to the fluid. In the case of the ocean the external pressure applied at its surface is the atmospheric pressure.

pressure at 10 you can determine the normal force on a unit area at that depth. Also recall that in general the pressure in a fluid varies with height but it also depends on the external pressure applied to the fluid. In the case of the ocean the external pressure applied at its surface is the atmospheric pressure. The weight of a column of seawater 1 in cross section and 10 high is about 2.00 10 5. The weight of a column of seawater 1 in cross section and 10 high plus the weight of a column of air with the same cross section extending up to the top of the atmosphere is about 2.00 10 5. The weight of 1 of seawater at 10 below the surface of the ocean is about 2.00 10 5. The density of seawater is about 2.00 10 5 times the density of air at sea level. The pressure at a given level in a fluid is caused by the weight of the overlying fluid plus any external pressure applied to the fluid. In this case the external pressure is the atmospheric pressure. Tropical storms are formed by low-pressure systems in the atmosphere. When such a storm forms over the ocean both the atmospheric pressure and the fluid pressure in the ocean decrease. Part B Now consider the pressure 20 is true? B.1 Variation of pressure with height below the surface of the ocean. Which of the following statements If the pressure at a point in a fluid is then at a distance below this point the pressure in the fluid is where is the density of the fluid and is the acceleration due to gravity. This means that the pressure at a depth below a reference height of pressure is greater than by an amount. Note that the quantity is related to the weight of a column of fluid of height and 1 in cross section. The pressure is twice that at a depth of 10. The pressure is the same as that at a depth of 10.

The pressure is equal to that at a depth of 10 plus the weight per 1 cross sectional area of a column of seawater 10 high. The pressure is equal to the weight per 1 column of seawater 20 high. cross sectional area of a Tactics Box 13.1 Hydrostatics Description: Knight/Jones/Field Tactics Box 13.1 Hydrostatics is illustrated. Learning Goal: To practice Tactics Box 13.1 Hydrostatics. In problems about liquids in hydrostatic equilibrium you often need to find the pressure at some point in the liquid. This Tactics Box outlines a set of rules for thinking about such hydrostatic problems. TACTICS BOX 13.1 Hydrostatics 1. Draw a picture. Show open surfaces pistons boundaries and other features that affect pressure. Include height and area measurements and fluid densities. Identify the points at which you need to find the pressure. These objects make up the system; the environment is everything else. 2. Determine the pressure at surfaces. Surface open to the air: usually 1. Surface in contact with a gas:. Closed surface: where is the force the surface such as a piston exerts on the fluid and is the area of the surface. 3. Use horizontal lines. The pressure in a connected fluid (of one kind) is the same at any point along a horizontal line. 4. Allow for gauge pressure. Pressure gauges read. 5. Use the hydrostatic pressure equation: where is the density of the fluid is the acceleration due to gravity and is the height of the fluid. Use these rules to work out the following problem: A U-shaped tube is connected to a box at one end and open to the air at the other end. The box is full of gas at pressure and the tube is filled with mercury of density 1.36 10 4. When the liquid in the tube reaches static equilibrium the mercury column is = 16.0 high in the left arm and = 6.00 high in the right arm as

shown in the figure. What is the gas pressure Part A inside the box? Find the pressures and at surfaces A and B in the tube respectively. Use to denote atmospheric pressure. Express your answers separated by a comma in terms of one or both of the variables. = and Part B As stated in rule 3 in the Tactics Box it is always convenient to use horizontal lines in hydrostatic problems. In each one of the following sketches a different horizontal line is considered. Which sketch would be more useful in solving the problem of finding the gas pressure?

It is useful to draw a horizontal line as in this sketch because it allows you to relate pressure in the mercury at critical points in the tube. For example even though you are not given the pressure at point C you know that because point C is at the same height as any point on surface B. At the same time you can also use the hydrostatic pressure equation to relate the pressure at point C to that at surface A where pressure is determined by the gas pressure in the box. Part C Assume. What is the gas pressure? C.1 Recall that Conversion factor between atmospheres and pascals. C.2 Find the height of the liquid column above point C What is the height of the mercury column above point C? Express your answer in meters. = Express your answer in pascals to three significant figures. =

= A Submerged Ball Description: Quantitative question involving the application of Archimedes' principle. A ball of mass and volume is lowered on a string into a fluid of density. would sink to the bottom if it were not supported by the string. Part A Assume that the object What is the tension shown in the figure? A.1 Equilibrium condition in the string when the ball is fully submerged but not touching the bottom as Although the fact may be obscured by the presence of a liquid the basic condition for equilibrium still holds: The net force on the ball must be zero. Draw a free-body diagram and proceed from there. A.2 Find the magnitude of the buoyant force Find A.2.1 the magnitude of the buoyant force. Archimedes' principle Quantitatively the buoyant force can be found as

A.2.1 Archimedes' principle Quantitatively the buoyant force can be found as where is the force is the density of the fluid is the magnitude of the acceleration due to gravity and is the volume of the displaced fluid. A.2.2 Find the mass of the displaced fluid Compute the mass of the fluid displaced by the object when it is entirely submerged. Express your answer in terms of any or all of the variables and. = Express your answer in terms of any or all of the variables and. = Express your answer in terms of any or all of the given quantities and acceleration due to gravity. the magnitude of the = ± Buoyant Force Conceptual Question Description: ± Includes Math Remediation. Conceptual question on the properties of a floating wooden block. A rectangular wooden block of weight floats with exactly one-half of its volume below the

waterline. Part A What is the buoyant force acting on the block? A.1 Archimedes' principle The upward buoyant force on a floating (or submerged) object is equal to the weight of the liquid displaced by the object. Mathematically the buoyant force on a floating (or submerged) object is where is the density of the fluid is the submerged volume of the object and is the acceleration due to gravity. A.2 What happens at equilibrium The block is in equilibrium so the net force acting on it is equal to zero. The buoyant force cannot be determined.

Part B The density of water is 1.00 B.1 Applying Archimedes' principle. What is the density of the block? In Part A you determined that the buoyant force and hence the weight of the water displaced is equal to the weight of the block. Notice however that the volume of the water displaced is onehalf of the volume of the block. B.2 Density The density of a material of mass and volume is 2.00. between 1.00 and 2.00 1.00 between 0.50 and 1.00 0.50 The density cannot be determined. Part C Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 C.1 of mass. What is the mass of the wooden block? How to approach the problem When you add the extra mass on top of the block the buoyant force must change. Relating the mass to the new buoyant force will allow you to write an equation to solve for. C.2 Find the new buoyant force With the 20- mass on the block twice as much of the block is underwater. Therefore what happens to the buoyant force on the block? The buoyant force doubles. The buoyant force is halved. The buoyant force doesn't change.

The buoyant force doubles. The buoyant force is halved. The buoyant force doesn't change. C.3 System mass Before the 20- mass is placed on the block the mass of the system is just the mass of the block ( ) and this mass is supported by the buoyant force. With the 20- mass on top the total mass of the system is now. C.4 The mass equation The original buoyant force was equal to the weight is twice the old buoyant force then of the block. If the new buoyant force where is the mass of the block. The new buoyant force must support the weight of the block and the mass so according to Newton's 2nd law By setting these two expressions for equal to each other you can solve for.. 40 20 10 Part D The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid only one-third of the wooden block is submerged. Is the unknown liquid more or less dense than water? more dense less dense Part E

What is the density of the unknown liquid? E.1 Comparing densities From Part C the mass of the block is 20. In water one-half of the block is submerged so onehalf of the volume of the block times the density of the water must be equivalent to 20. In the unknown substance one-third of the block is submerged so one-third of the volume of the block times the density of the unknown substance must be equivalent to 20. You can use these observations to write an equation for the density of the unknown substance. E.2 Setting up the equation The buoyant force on the block in water is the same as the buoyant force in the unknown liquid so where is the density of water is the density of the unknown liquid is the volume of the entire block and is the acceleration due to gravity. Express your answer numerically in grams per cubic centimeter. = Understanding Bernoulli's Equation Description: Conceptual questions on Bernoulli's equation applied first to a horizontal converging flow tube and then to an elevated one. Bernoulli's equation is a simple relation that can give useful insight into the balance among fluid pressure flow speed and elevation. It applies exclusively to ideal fluids with steady flow that is fluids with a constant density and no internal friction forces whose flow patterns do not change with time. Despite its limitations however Bernoulli's equation is an essential tool in understanding the behavior of fluids in many practical applications from plumbing systems to the flight of airplanes. For a fluid element of density that flows along a streamline Bernoulli's equation states that where is the pressure is the flow speed is the height is the acceleration due to gravity and

subscripts 1 and 2 refer to any two points along the streamline. The physical interpretation of Bernoulli's equation becomes clearer if we rearrange the terms of the equation as follows: The term on the left-hand side represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from point 1 to point 2. The two terms on the right-hand side represent respectively the change in potential energy. and the change in kinetic energy of the unit volume during its flow from point 1 to point 2. In other words Bernoulli's equation states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occurs during the flow. This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing along a streamline. Part A Consider the portion of a flow tube shown in the figure. Point 1 and point 2 are at the same height. An ideal fluid enters the flow tube at point 1 and moves steadily toward point 2. If the cross section of the flow tube at point 1 is greater than that at point 2 what can you say about the pressure at point 2? A.1 How to approach the problem Apply Bernoulli's equation to point 1 and to point 2. Since the points are both at the same height their elevations cancel out in the equation and you are left with a relation between pressure and flow speeds. Even though the problem does not give direct information on the flow speed along the flow tube it does tell you that the cross section of the flow tube decreases as the fluid flows toward point 2. Apply the continuity equation to points 1 and 2 and determine whether the flow speed at point 2 is greater than or smaller than the flow speed at point 1. With that information and Bernoulli's equation you will be able to determine the pressure at point 2 with respect to the pressure at point 1. A.2 Apply Bernoulli's equation Apply Bernoulli's equation to point 1 and to point 2 to complete the expression below. Here and

point 2 is greater than or smaller than the flow speed at point 1. With that information and Bernoulli's equation you will be able to determine the pressure at point 2 with respect to the pressure at point 1. A.2 Apply Bernoulli's equation Apply Bernoulli's equation to point 1 and to point 2 to complete the expression below. Here are the pressure and flow speed respectively and subscripts 1 and 2 refer to point 1 and point 2. Also use for elevation with the appropriate subscript and use for the density of the fluid. A.2.1 Flow along a horizontal streamline Along a horizontal streamline the change in potential energy of the flowing fluid is zero. In other words when applying Bernoulli's equation to any two points of the streamline and they cancel out. and Express your answer in terms of some or all of the variables and. = Along a horizontal streamline Bernoulli's equation reduces to or. Since the density of the fluid is always a positive quantity the sign of gives you the sign of allowing you to determine whether is greater than or smaller than. A.3 Determine with respect to By applying the continuity equation determine which of the following is true. A.3.1 The continuity equation The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube. It says that the amount of fluid flowing through a cross section of the tube in a time interval must be the same for all cross sections or. Therefore the flow speed must increase when the cross section of the flow tube decreases and vice versa.

time interval must be the same for all cross sections or. Therefore the flow speed must increase when the cross section of the flow tube decreases and vice versa. The pressure at point 2 is lower than the pressure at point 1. equal to the pressure at point 1. higher than the pressure at point 1. Thus by combining the continuity equation and Bernoulli's equation one can characterize the flow of an ideal fluid.when the cross section of the flow tube decreases the flow speed increases and therefore the pressure decreases. In other words if then and. Part B As you found out in the previous part Bernoulli's equation tells us that a fluid element that flows through a flow tube with decreasing cross section moves toward a region of lower pressure. Physically the pressure drop experienced by the fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to. B.1 Effects from conservation of mass Recall that if the cross section of the flow tube varies the flow speed must change to conserve mass. This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decrease speed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger cross section. decrease in speed increase in speed remain in equilibrium Part C Now assume that point 2 is at height with respect to point 1 as shown in the figure.

The ends of the flow tube have the same areas as the ends of the horizontal flow tube shown in Part A. Since the cross section of the flow tube is decreasing Bernoulli's equation tells us that a fluid element flowing toward point 2 from point 1 moves toward a region of lower pressure. In this case what is the pressure drop experienced by the fluid element? C.1 How to approach the problem Apply Bernoulli's equation to point 1 and to point 2 as you did in Part A. Note that this time you must take into account the difference in elevation between points 1 and 2. Do you need to add this additional term to the other term representing the pressure drop between the two ends of the flow tube or do you subtract it? The pressure drop is smaller than the pressure drop occurring in a purely horizontal flow. equal to the pressure drop occurring in a purely horizontal flow. larger than the pressure drop occurring in a purely horizontal flow. Part D From a physical point of view how do you explain the fact that the pressure drop at the ends of the elevated flow tube from Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A? D.1 Physical meaning of the pressure drop in a tube As explained in the introduction the difference in pressure between the ends of a flow tube represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from one end to the other end of the flow tube. A greater amount of work is needed to balance the increase in potential energy from the elevation change. decrease in potential energy from the elevation change. larger increase in kinetic energy.

surrounding fluid to move that volume of fluid from one end to the other end of the flow tube. A greater amount of work is needed to balance the increase in potential energy from the elevation change. decrease in potential energy from the elevation change. larger increase in kinetic energy. larger decrease in kinetic energy. In the case of purely horizontal flow the difference in pressure between the two ends of the flow tube had to balance only the increase in kinetic energy resulting from the acceleration of the fluid. In an elevated flow tube the difference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure is needed for the flow to occur. Flow Velocity of Blood Conceptual Question Description: Conceptual question on plaque changing the effective cross sectional area and blood pressure of an artery. Arteriosclerotic plaques forming on the inner walls of arteries can decrease the effective cross-sectional area of an artery. Even small changes in the effective area of an artery can lead to very large changes in the blood pressure in the artery and possibly to the collapse of the blood vessel. Imagine a healthy artery with blood flow velocity of and mass per unit volume of. The kinetic energy per unit volume of blood is given by Imagine that plaque has narrowed an artery to one-fifth of its normal cross-sectional area (an 80% blockage). Part A Compared to normal blood flow velocity blockage? A.1 what is the velocity of blood as it passes through this Continuitity equation and reduced cross-sectional area By the equation of continuity as the cross-sectional area of an artery decreases because of plaque formation the velocity of blood through that region of the artery will increase. The new flow speed can be calculated by rearranging the equation of continuity so ; where and are the initial and final cross-sectional areas and and are the initial and final velocities of the blood respectively.

Part B By what factor does the kinetic energy per unit of blood volume change as the blood passes through this blockage? 25 5 1 Part C As the blood passes through this blockage what happens to the blood pressure? C.1 Blood pressure and blood velocity Bernoulli's equation states that the sum of the pressure the kinetic energy per volume and the gravitational energy per volume of a fluid is constant. For initial and final pressures and initial and final velocities and and mass per unit volume of blood ignoring the effects of changes in gravitational energy leads to. Basically the sum of kinetic energy and pressure must remain constant in an artery. This leads to a very serious health risk. As blood velocity increases blood pressure in a section of artery can drop to a dangerously low level and the blood vessel can collapse completely cutting off blood flow owing to lack of sufficient internal pressure. C.2 Calculating the change in blood pressure From Bernoulli's equation the change in pressure is the negative of the change in kinetic energy per unit volume. For initial and final kinetic energies per unit volume of the blood and respectively or where. Rearranging this equation yields

or where. Rearranging this equation yields or. It increases by about 250. It increases by about 41. It stays the same. It decreases by about 41. It decreases by about 250. Since the kinetic energy increases by a factor of 25 Bernoulli's equation tells you that. As the blood velocity increases through a blockage the blood pressure in that section of the artery can drop to a dangerously low level. In extreme cases the blood vessel can collapse completely cutting off blood flow owing to lack of sufficient internal pressure. In the next three parts you will see how a small increase in blockage can cause a much larger pressure change. For parts D - F imagine that plaque has grown to a 90% blockage.. Part D Relative to its initial healthy state by what factor does the velocity of blood increase as the blood passes through this blockage? Express your answer numerically. Part E By what factor does the kinetic energy per unit of blood volume increase as the blood passes through this blockage? Express your answer numerically.

Express your answer numerically. Part F What is the magnitude of the drop in blood pressure as the blood passes through this blockage? Use as the normal (i.e. unblocked) kinetic energy per unit volume of the blood. Express your answer in pascals using two significant figures. = Video Tutor: Air Jet Blows between Bowling Balls Description: An air jet is directed between a pair of bowling balls hanging close to each other. How do the balls respond? First launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then close the video window and answer the question on the right. You can watch the video again at any point. Part A A strong wind blows over the house shown in the figure. The wind is much stronger over the house's roof than lower down and the house has an open chimney. A window on the ground floor is open and so are the doors inside the house. Which way will air flow through the house?

A.1 How to approach the problem This problem asks you to apply what you've learned about pressure and fluid flow to a new situation. Is the pressure above the chimney lower or higher than that at the window? Remember that the wind is much faster at the level of the chimney than at the level of the open window. Wind will flow from a region of higher pressure to a region of lower pressure. Use this idea to determine the direction of air flow. In the window and out the chimney In no direction; there is no air flow through the house. In the chimney and out the window Air pressure is lower at the chimney and higher outside the window so air will flow in through the window and out through the chimney. Video Tutor: Pressure in Water and Alcohol Description: Pressure is measured at the bottom of two equal-height columns of liquid one consisting of water and the other of alcohol. How do the pressures compare? First launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then close the video window and answer the question on the right. You can watch the video again at any point.

Part A Rank from smallest to largest the pressures in the tank of motionless fluid shown in the figure. A.1 How to approach the problem Recall that the pressure at any point in an open container of fluid is due to the weight of the overlying fluid plus the weight of the overlying air. Also think about the fact that this fluid is motionless. What does that tell you about the pressure at any two points on the same horizontal level such as points B and C? To rank items as equivalent overlap them. Pressure depends on vertical distance from the fluid surface. As we descend deeper in the fluid the pressure increases.

a cylinder is = so if we double the radius we must reduce the depth of immersion by a View Pressure depends on vertical distance from the fluid surface. As we descend deeper in the fluid the pressure increases. Video Tutor: Weighing Weights in Water Description: Steel and aluminum cylinders of identical size and shape are attached to spring scales and lowered into water. Which if either shows a greater reduction in weight? First launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then close the video window and answer the question on the right. You can watch the video again at any point. Part A Suppose that we repeat the experiment shown in the video but we replace one of the cylinders with a cylinder that has twice the radius (and use larger containers of water). If the height of the original cylinder is the video? A.1 how deeply must we submerge the new cylinder to get the same weight reduction as in How to approach the problem How does the reduction in the cylinder s weight relate to the amount of water the cylinder displaces? How does the volume of a cylinder depend on the cylinder s radius and height? 2 To get the same reduction in weight the same volume of cylinder must be submerged. The volume of

4 To get the same reduction in weight the same volume of cylinder must be submerged. The volume of a cylinder is = so if we double the radius we must reduce the depth of immersion by a factor of four. Q13.33. Reason: Since the question talks about the extra pressure we will ignore the air pressure above the water; you have an equal amount of air pressure on the inside. The 7 N quoted is the increased pressure. Therefore the equation we need is Equation 13.5 without the p o term where we want to solve for We will also use and round g to for one significant figure accuracy. The correct answer is D. Assess: This one significant figure calculation can easily be done in the head without a calculator. The answer seems plausible and the other choices seem too small. The units cancel appropriately to leave the answer in m. Q13.35. Reason: We ll assume the ball is not being forcibly held under the water (because then the net force would be zero). A free-body diagram shows only two forces on the ball: the downward force of gravity and the upward buoyant force whose magnitude equals the weight of the displaced fluid (the weight of of water). The downward gravitational force has a magnitude of The upward buoyant force has a magnitude of Subtracting gives the magnitude of the net force The correct answer is B. Assess: This is enough force to cause the basketball to accelerate upward fairly impressively as you have probably seen in the swimming pool (if you haven t have your teacher assign you a field trip to the pool to try this out). Q13.37. Reason: We ll use the equation of continuity Equation 13.12 and make all the assumptions inherent in that equation (no leakage etc.). We ll also use Call the plunger point 1 and the nozzle point 2 and solve for v 1.

The correct answer is B. Assess: The answer seems to be a reasonable slow and steady pace. The ratio of the speeds is the square of the ratio of the radii. The last step (the computation) can easily be done without a calculator. Q13.38. Reason: The question says to ignore viscosity so we do not need Poiseuille s equation; Bernoulli s equation should suffice. And because the pipe is horizontal we can drop the terms (because they will be the same on both sides). We are given and We will also use the equation of continuity to solve for v 2. Putting it all together The magnitude of this is 12700 Pa so the correct answer is D. Assess: Since Pa = N/m 2 the units work out. Problems P13.7. Prepare: V f is the volume of the fish V a is the volume of air and V w is the volume of water displaced by the fish + air system. The mass of the fish is m f the mass of the air is m a and the mass of the water displaced by the fish is m w. Solve: Neutral buoyancy implies static equilibrium which will happen when This means Assess: Increase in volume of 8.0% seems very reasonable. P13.8. Prepare: The density of sea water is 1030 kg/m 3. Also Solve: The pressure below sea level can be found from Equation 13.5 as follows: Assess: The pressure deep in the ocean is very large.

P13.14. Prepare: The pressure required will be the atmospheric pressure exerted by the air on the water in the container subtracted by the pressure exerted by the column of water. Solve: Assess: The principle here is similar to the principle in a barometer. P13.30. Prepare: The pipe is a flow tube so the equation of continuity Equation 13.12 applies. To work with SI units we need the conversion 1 L = 10 3 m 3. Solve: The volume flow rate is Using the definition Q = va we get Assess: This is a reasonable speed for water flowing through a 2.0-cm pipe. P13.55. Prepare: Knowing the volume flow rate the number of vessels and the diameter of each vessel we can use Equation 13.13 to solve the problem. We can either say that each vessel handles 1/2000 of the volume flow rate or we can say the total area for this flow rate is 2000 times the area of each vessel. Either way we will get the same result. Solve: Starting with the definition of volume flow rate and the knowledge that each vessel handles 1/2000 of this volume we may write Solving for v Assess: At this rate water could travel to the top of an 18 m tree in a day. That seems reasonable. P13.61. Prepare: Please refer to Figure P13.61. Assume the gas in the manometer is an ideal gas. Assume that a drop in length of 90 mm produces a very small change in gas volume compared with the total volume of the gas cell. This means the volume of the chamber can be considered constant. Solve: In the ice-water mixture the pressure is p 1 = p atmos + ρ Hg g (0.120 m) = 1.013 10 5 Pa + (13600 kg/m 3 )(9.8 m/s 2 )(0.120 m) = 1.173 10 5 Pa In the freezer the pressure is p 2 = p atm + ρ Hg g (0.030 m) = 1.013 10 5 Pa + (13600 kg/m 3 )(9.8 m/s 2 )(0.030 m) = 1.053 10 5 Pa For constant volume conditions the gas equation is Assess: This is a reasonable temperature for an industrial freezer.